51–52MechanicsCore

Viscosity and Terminal Velocity

Honey pours slowly and water pours quickly because honey is more viscous\textbf{viscous} — it resists flowing. A small sphere falling through a viscous liquid quickly stops accelerating and sinks at a steady terminal velocity\textbf{terminal velocity}, and measuring that speed is how viscosity is found in the lab.

35 min Topic 4 · Materials

What you'll be able to do

  • Describe viscosity and distinguish laminar from turbulent flow
  • Use Stokes’ law F=6πηrvF = 6\pi\eta r v and state its conditions
  • Explain terminal velocity in terms of the forces on a falling object
  • Derive and use the expression for terminal velocity of a sphere
1

Viscosity and types of flow

Viscosity measures a fluid’s internal resistance to flow — how strongly adjacent layers drag on each other. It is quantified by the coefficient of viscosity\textbf{coefficient of viscosity} η\eta, measured in Pa s\mathrm{Pa\,s}.

In laminar\textbf{laminar} flow the fluid moves in smooth parallel layers, with no mixing between them and velocity at any point constant over time. In turbulent\textbf{turbulent} flow the motion becomes chaotic, with eddies and swirls, and the drag rises sharply.

Viscosity is strongly temperature dependent. Most liquids become much less viscous when warmed — which is why engine oil flows more freely once warm, and why temperature must be controlled in the viscosity practical.

[η]=Pa s=kg m−1 s−1[\eta] = \mathrm{Pa\,s} = \mathrm{kg\,m^{-1}\,s^{-1}}
Units of viscosity.
2

Stokes’ law

For a small sphere moving slowly through a fluid in laminar flow, the viscous drag is F=6πηrvF = 6\pi\eta r v, where rr is the sphere’s radius and vv its speed relative to the fluid.

The drag is proportional to speed. That linearity is what makes terminal velocity so easy to reach and to analyse: as the sphere speeds up, the drag grows in step until it balances the other forces.

The law only holds for small spheres, low speeds and laminar flow. It fails for large or fast objects, where turbulence makes drag roughly proportional to v2v^{2} instead.

F=6πηrvF = 6\pi\eta r v
Stokes’ law — laminar flow only.
Example 1A sphere of radius 1.5 mm moves at 0.020 m s−10.020\ \mathrm{m\,s^{-1}} through glycerol with η=1.4 Pa s\eta = 1.4\ \mathrm{Pa\,s}. Find the viscous drag.
1r=1.5×10−3r = 1.5\times10^{-3} m.
2F=6π×1.4×1.5×10−3×0.020F = 6\pi \times 1.4 \times 1.5\times10^{-3} \times 0.020.
AnswerF=7.9×10−4 NF = 7.9\times10^{-4}\ \mathrm{N}

Tip — Use the radius, not the diameter. Micrometer readings give diameter, and halving it is easily forgotten.

3

Reaching terminal velocity

A sphere released in a liquid has three forces on it: weight downwards, upthrust upwards, and viscous drag upwards (opposing its downward motion).

At release the speed is zero, so drag is zero and the resultant force is weight minus upthrust — the sphere accelerates. As it speeds up, drag increases, the resultant falls, and so does the acceleration.

Eventually drag grows until weight equals upthrust plus drag. The resultant force is then zero, and by Newton’s first law the sphere continues at constant speed: its terminal velocity\textbf{terminal velocity}.

W=U+F at terminal velocityW = U + F \text{ at terminal velocity}
Forces balance.
4

An expression for terminal velocity

Substituting each force for a sphere of density ρs\rho_{s} in a fluid of density ρf\rho_{f}: weight 43πr3ρsg\tfrac43\pi r^{3}\rho_{s}g, upthrust 43πr3ρfg\tfrac43\pi r^{3}\rho_{f}g, drag 6πηrvt6\pi\eta r v_{t}.

Setting weight equal to upthrust plus drag and rearranging gives the terminal velocity. It grows with r2r^{2} — so doubling the radius quadruples the terminal speed — and falls with increasing viscosity.

In the practical, measuring vtv_{t} with light gates or by timing between markers lets you calculate η\eta. Markers are placed far enough down the tube that the ball has already reached terminal velocity before timing starts.

vt=2r2g(ρs−ρf)9ηv_{t} = \dfrac{2r^{2}g\left(\rho_{s}-\rho_{f}\right)}{9\eta}
Terminal velocity of a sphere in laminar flow.
Example 1A steel ball (ρs=7800 kg m−3\rho_{s} = 7800\ \mathrm{kg\,m^{-3}}) of radius 1.0 mm falls through oil (ρf=900 kg m−3\rho_{f} = 900\ \mathrm{kg\,m^{-3}}) at a terminal velocity of 0.050 m s−10.050\ \mathrm{m\,s^{-1}}. Find η\eta. Take g=9.81g = 9.81.
1Rearrange: η=2r2g(ρs−ρf)9vt\eta = \dfrac{2r^{2}g(\rho_{s}-\rho_{f})}{9v_{t}}.
2Numerator: 2×(1.0×10−3)2×9.81×6900=0.135382 \times (1.0\times10^{-3})^{2} \times 9.81 \times 6900 = 0.13538.
3Denominator: 9×0.050=0.459 \times 0.050 = 0.45.
Answerη=0.30 Pa s\eta = 0.30\ \mathrm{Pa\,s}

Tip — Timing must start only after terminal velocity is reached. Placing the first marker too near the surface measures an average speed that is too low, overestimating viscosity.

Equation recap

F=6πηrvF = 6\pi\eta rv
Stokes’ law.
W=U+FW = U + F
At terminal velocity.
vt=2r2g(ρs−ρf)9ηv_{t} = \tfrac{2r^{2}g(\rho_{s}-\rho_{f})}{9\eta}
Terminal velocity.
vt∝r2v_{t} \propto r^{2}
Doubling radius quadruples speed.

Common mistakes to avoid

Omitting upthrust at terminal velocity.
Weight balances upthrust plus\textit{plus} drag.
Using diameter in Stokes’ law.
Use the radius.
Applying Stokes’ law to a skydiver.
That flow is turbulent; Stokes’ law is only for slow laminar flow.
Saying no forces act at terminal velocity.
The forces are balanced, so the resultant\textit{resultant} is zero.

Key takeaways

  • Viscosity η\eta measures resistance to flow, in Pa s\mathrm{Pa\,s}, and falls with temperature.
  • Stokes’ law F=6πηrvF = 6\pi\eta rv applies to small spheres in slow laminar flow.
  • A falling sphere accelerates until weight equals upthrust plus drag.
  • vt=2r2g(ρs−ρf)/9ηv_{t} = 2r^{2}g(\rho_{s}-\rho_{f})/9\eta, so terminal speed scales with r2r^{2}.
  • In the practical, time only after terminal velocity is reached.

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